Integrand size = 21, antiderivative size = 123 \[ \int (a+a \sec (c+d x))^n \tan ^7(c+d x) \, dx=\frac {7 (a+a \sec (c+d x))^{4+n}}{a^4 d (4+n)}+\frac {\operatorname {Hypergeometric2F1}(1,4+n,5+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{4+n}}{a^4 d (4+n)}-\frac {5 (a+a \sec (c+d x))^{5+n}}{a^5 d (5+n)}+\frac {(a+a \sec (c+d x))^{6+n}}{a^6 d (6+n)} \] Output:
7*(a+a*sec(d*x+c))^(4+n)/a^4/d/(4+n)+hypergeom([1, 4+n],[5+n],1+sec(d*x+c) )*(a+a*sec(d*x+c))^(4+n)/a^4/d/(4+n)-5*(a+a*sec(d*x+c))^(5+n)/a^5/d/(5+n)+ (a+a*sec(d*x+c))^(6+n)/a^6/d/(6+n)
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71 \[ \int (a+a \sec (c+d x))^n \tan ^7(c+d x) \, dx=\frac {(1+\sec (c+d x))^4 (a (1+\sec (c+d x)))^n \left (\frac {7}{4+n}+\frac {\operatorname {Hypergeometric2F1}(1,4+n,5+n,1+\sec (c+d x))}{4+n}-\frac {5 (1+\sec (c+d x))}{5+n}+\frac {(1+\sec (c+d x))^2}{6+n}\right )}{d} \] Input:
Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^7,x]
Output:
((1 + Sec[c + d*x])^4*(a*(1 + Sec[c + d*x]))^n*(7/(4 + n) + Hypergeometric 2F1[1, 4 + n, 5 + n, 1 + Sec[c + d*x]]/(4 + n) - (5*(1 + Sec[c + d*x]))/(5 + n) + (1 + Sec[c + d*x])^2/(6 + n)))/d
Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 25, 4368, 25, 27, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \tan ^7(c+d x) (a \sec (c+d x)+a)^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^7 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^ndx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^7 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^ndx\) |
\(\Big \downarrow \) 4368 |
\(\displaystyle \frac {\int -a^3 \cos (c+d x) (1-\sec (c+d x))^3 (\sec (c+d x) a+a)^{n+3}d\sec (c+d x)}{a^6 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int a^3 \cos (c+d x) (1-\sec (c+d x))^3 (\sec (c+d x) a+a)^{n+3}d\sec (c+d x)}{a^6 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \cos (c+d x) (1-\sec (c+d x))^3 (\sec (c+d x) a+a)^{n+3}d\sec (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (\cos (c+d x) (\sec (c+d x) a+a)^{n+3}-7 (\sec (c+d x) a+a)^{n+3}+\frac {5 (\sec (c+d x) a+a)^{n+4}}{a}-\frac {(\sec (c+d x) a+a)^{n+5}}{a^2}\right )d\sec (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {(a \sec (c+d x)+a)^{n+6}}{a^3 (n+6)}+\frac {5 (a \sec (c+d x)+a)^{n+5}}{a^2 (n+5)}-\frac {(a \sec (c+d x)+a)^{n+4} \operatorname {Hypergeometric2F1}(1,n+4,n+5,\sec (c+d x)+1)}{a (n+4)}-\frac {7 (a \sec (c+d x)+a)^{n+4}}{a (n+4)}}{a^3 d}\) |
Input:
Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^7,x]
Output:
-(((-7*(a + a*Sec[c + d*x])^(4 + n))/(a*(4 + n)) - (Hypergeometric2F1[1, 4 + n, 5 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(4 + n))/(a*(4 + n)) + (5*(a + a*Sec[c + d*x])^(5 + n))/(a^2*(5 + n)) - (a + a*Sec[c + d*x])^(6 + n)/(a^3*(6 + n)))/(a^3*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(d*b^(m - 1))^(-1) Subst[Int[(-a + b*x)^((m - 1)/2 )*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a, b, c , d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && !IntegerQ[n]
\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{7}d x\]
Input:
int((a+a*sec(d*x+c))^n*tan(d*x+c)^7,x)
Output:
int((a+a*sec(d*x+c))^n*tan(d*x+c)^7,x)
\[ \int (a+a \sec (c+d x))^n \tan ^7(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{7} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^7,x, algorithm="fricas")
Output:
integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^7, x)
Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^7(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**n*tan(d*x+c)**7,x)
Output:
Timed out
\[ \int (a+a \sec (c+d x))^n \tan ^7(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{7} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^7,x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^7, x)
\[ \int (a+a \sec (c+d x))^n \tan ^7(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{7} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^7,x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^7, x)
Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^7(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^7\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \] Input:
int(tan(c + d*x)^7*(a + a/cos(c + d*x))^n,x)
Output:
int(tan(c + d*x)^7*(a + a/cos(c + d*x))^n, x)
\[ \int (a+a \sec (c+d x))^n \tan ^7(c+d x) \, dx=\text {too large to display} \] Input:
int((a+a*sec(d*x+c))^n*tan(d*x+c)^7,x)
Output:
((sec(c + d*x)*a + a)**n*tan(c + d*x)**6*n**3 + 6*(sec(c + d*x)*a + a)**n* tan(c + d*x)**6*n**2 + 8*(sec(c + d*x)*a + a)**n*tan(c + d*x)**6*n - 6*(se c(c + d*x)*a + a)**n*tan(c + d*x)**4*n**2 - 12*(sec(c + d*x)*a + a)**n*tan (c + d*x)**4*n + 24*(sec(c + d*x)*a + a)**n*tan(c + d*x)**2*n - 48*(sec(c + d*x)*a + a)**n + int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**7)/(sec(c + d*x)*n**3 + 12*sec(c + d*x)*n**2 + 44*sec(c + d*x)*n + 48*sec(c + d*x) + n **3 + 12*n**2 + 44*n + 48),x)*d*n**7 + 18*int(((sec(c + d*x)*a + a)**n*tan (c + d*x)**7)/(sec(c + d*x)*n**3 + 12*sec(c + d*x)*n**2 + 44*sec(c + d*x)* n + 48*sec(c + d*x) + n**3 + 12*n**2 + 44*n + 48),x)*d*n**6 + 124*int(((se c(c + d*x)*a + a)**n*tan(c + d*x)**7)/(sec(c + d*x)*n**3 + 12*sec(c + d*x) *n**2 + 44*sec(c + d*x)*n + 48*sec(c + d*x) + n**3 + 12*n**2 + 44*n + 48), x)*d*n**5 + 408*int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**7)/(sec(c + d*x )*n**3 + 12*sec(c + d*x)*n**2 + 44*sec(c + d*x)*n + 48*sec(c + d*x) + n**3 + 12*n**2 + 44*n + 48),x)*d*n**4 + 640*int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**7)/(sec(c + d*x)*n**3 + 12*sec(c + d*x)*n**2 + 44*sec(c + d*x)*n + 48*sec(c + d*x) + n**3 + 12*n**2 + 44*n + 48),x)*d*n**3 + 384*int(((sec( c + d*x)*a + a)**n*tan(c + d*x)**7)/(sec(c + d*x)*n**3 + 12*sec(c + d*x)*n **2 + 44*sec(c + d*x)*n + 48*sec(c + d*x) + n**3 + 12*n**2 + 44*n + 48),x) *d*n**2 - 6*int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**5)/(sec(c + d*x)*n* *3 + 12*sec(c + d*x)*n**2 + 44*sec(c + d*x)*n + 48*sec(c + d*x) + n**3 ...